Maths problem
回答 (2)
2006-10-16 11:27 pm
✔ 最佳答案
ax^2+bx+c=0= x^2+(b/a)x + c/a=0
(x+ b/2a)^2 - b^2/4a^2+c/a=0
(x+b/2a)= b^2/4a^2-c/a
(x+b/2a)=(b^2-4ac)/4a^2
(x+b/2a)= + (√(b^2-4ac))/2a or - (√(b^2-4ac))/2a
therefore x = [-b + (√(b^2-4ac)]/2a or [-b - (√(b^2-4ac)]/2a
2006-10-16 11:28 pm
ax^2 + bx + c = 0
a(x^2+b/a x + (b/2a)^2 - (b/2a)^2)+c = 0
a(x+b/2a)^2 - b^2/4a + c = 0
(x+b/2a)^2 = b^2/4a^2 - c/a
x = -b/2a +/- ( b^2/4a^2 - c/a)^2
x = (-b +/- ( b^2 - 4ac)^2)/2a
a(x^2+b/a x + (b/2a)^2 - (b/2a)^2)+c = 0
a(x+b/2a)^2 - b^2/4a + c = 0
(x+b/2a)^2 = b^2/4a^2 - c/a
x = -b/2a +/- ( b^2/4a^2 - c/a)^2
x = (-b +/- ( b^2 - 4ac)^2)/2a
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